Is There A Better Way To Write || In A Single If Statment

was wondering if there is a better way to write the code below. I want my if statement to ignore all of the keycodes. but seems quite a messy way to write it as below. thanks! if (

Solution 1:

Using includes:

if ([9, 91, 20, 18, 17, 37].includes(event.keyCode))

Solution 2:

For a larger list of values that you compare with often, it may be more optimal to use a Set.

const codes = new Set([9, 91, 20, 18, 17, 37]);
if(codes.has(event.keyCode)){
   // do something
}

Solution 3:

You can use Array#some or Array#indexOf instead.

if ([9, 91, 20, 18, 17, 37].some(val => val === event.keyCode))

or

if ([9, 91, 20, 18, 17, 37].indexOf(event.keyCode) !== -1)

---

In terms of performance, I would highly recommend you should use indexOf.

• Array indexOf() vs includes() perfomance depending on browser and needle position

And another aspect you should take care of is between includes and indexOf in case of NAN issue. There are 2 posts below that can help you have comprehensive knowledge about them:

• Array.prototype.includes vs. Array.prototype.indexOf

• Why does [NaN].includes(NaN) return true in JavaScript?

Solution 4:

Like others have said, you can create an array or set with the valid key codes and test for membership using the event.keyCode variable. I wrote some example code below for how you could do this in a clean way :)

// let eventKeyCode = event.keyCode;let eventKeyCode = 28;
let validKeyCodes = [9, 91, 20, 18, 17, 37];
if (validKeyCodes.includes(eventKeyCode)) {
  // do something console.log(`event has valid key code: ${eventKeyCode}`);
} else {
  // catch logicconsole.log(`event has invalid key code: ${eventKeyCode}`);
}

Solution 5:

You can pass the if logic into a function.

functionisValidKey(keyCode) {
  const validKeys = [9, 91, 20, 18, 17, 37];
  return validKeys.includes(keyCode);
}
//...if (isValidKey(event.keyCode)) ...